Question: You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 15,\enspace 5,\enspace 9,\enspace 11,\enspace 9$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{15 + 5 + 9 + 11 + 9}{{5}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $15$ years $5.2$ years $27.04$ years $^2$ $5$ years $-4.8$ years $23.04$ years $^2$ $9$ years $-0.8$ years $0.64$ years $^2$ $11$ years $1.2$ years $1.44$ years $^2$ $9$ years $-0.8$ years $0.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{27.04} + {23.04} + {0.64} + {1.44} + {0.64}} {{5}} $ $ {\sigma^2} = \dfrac{{52.8}}{{5}} = {10.56\text{ years}^2} $ The average porcupine at the zoo is 9.8 years old. The population variance is 10.56 years $^2$.